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Tema [17]
3 years ago
9

11. What is dynamic equilibrium?​

Biology
2 answers:
valentina_108 [34]3 years ago
6 0

Answer:

Explanation:

Chemical reactions can either go in both directions (forward and reverse) or only in one direction. The ones that go in two directions are known as reversible reactions, and you can identify them by the arrows going in two directions, like the example below. 

H2O(l) ⇌ H+(aq) + OH-(aq)

Dynamic equilibrium only occurs in reversible reactions, and it’s when the rate of the forward reaction is equal to the rate of the reverse reaction. These equations are dynamic because the forward and reverse reactions are still occurring, but the two rates are equal and unchanging, so they’re also at equilibrium.

Dynamic equilibrium is an example of a system in a steady state. This means the variables in the equation are unchanging over time (since the rates of reaction are equal). If you look at a reaction in dynamic equilibrium, it’ll look like nothing is happening since the concentrations of each substance stay constant. However, reactions are actually continuously occurring.

--

Please give me brilliant

AlexFokin [52]3 years ago
6 0
Dynamic equilibrium. ... In chemistry, and in physics, a dynamic equilibrium exists once a reversible reaction occurs. Substances transition between the reactants and products at equal rates, meaning there is no net change. Reactants and products are formed at such a rate that the concentration of neither changes.
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3 years ago
g If a person takes a prescribed dose of 10 milligrams of Valium, the amount of Valium in that person's bloodstream at any time
Elenna [48]

Answer:

a. 8.1 milligrams

b. 40.07 hours

c. 8.859 milligrams

Explanation:

If a person takes a prescribed dose of 10 milligrams of Valium, the amount of Valium in that person's bloodstream at any time can be modeled by

A_{t}=10e^{-0.0173t}

Where A(t) = amount of Valium remaining in the blood after t hours

t = time or duration after the drug is taken

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A_{12}=10e^{-0.0173\times12}

A_{12}=10e^{-0.2076}

                = 10×0.81253

                = 8.1 milligrams

b. In this part we have to calculate the time when A(t) = 5 milligrams

5=10e^{-0.0173\timest}

\frac{5}{10}=e^{-0.0173t}

0.5 = e^{-0.0173t}

Now we take natural log on both the sides of the equation.

ln(0.5) = ln(e^{-0.0173t})

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t = \frac{0.69314}{0.0173}

t = 40.0658

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c. In this part we have to calculate the rate, by which amount of drug will decay in the bloodstream after 7 hours.

A_{7}=10e^{-0.0173\times7}

A_{7}=10e^{-0.1211}

              = 10×0.8859

              = 8.859 milligrams

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